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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Physics
Chapter
Electromagnetic Waves

JEE Main 2024 · 6 April · Evening Shift · Physics · Electromagnetic Waves

In the given electromagnetic wave E_ y =600 (ω t-kx) Vm^ -1 ,...

In the given electromagnetic wave Ey=600sin⁡(ωt−kx)Vm−1\mathrm{E}_{\mathrm{y}}=600 \sin (\omega \mathrm{t}-\mathrm{kx}) \mathrm{Vm}^{-1}, intensity of the associated light beam is (in W/m2\mathrm{W} / \mathrm{m}^2 : (Given ϵ0=9×10−12C2 N−1 m−2\epsilon_0=9 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2} )

Options

  1. A243
  2. B729
  3. C972
  4. D486

Correct answer

D. 486

Step-by-step solution

 Intensity =12ε0E02c=12×9×10−12×(600)2×3×108=92×36×3=486w/m2\begin{aligned} \text{ Intensity } & =\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \mathrm{c} \\ & =\frac{1}{2} \times 9 \times 10^{-12} \times(600)^2 \times 3 \times 10^8 \\ & =\frac{9}{2} \times 36 \times 3=486 \mathrm{w} / \mathrm{m}^2\end{aligned}
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